What is the integral of sec^8(x) ?

The integral of sec^8(x) is tan(x)+tan^3(x)+(3tan^5(x))/5+(tan^7(x))/7+C

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integral of sec^8(x)

\int sec^{8} (x) d x

tan (x) + tan^{3} (x) + \frac{3 tan ^{5} ( x )}{5} + \frac{tan ^{7} ( x )}{7} + C

Solution steps

\int sec^{8} (x) d x

Rewrite using trig identities

= \int sec^{2} (x) (1 + tan^{2} (x))^{3} d x

Apply u-substitution

: \int (1 + u^{2})^{3} d u

= \int (1 + u^{2})^{3} d u

Expand

(1 + u^{2})^{3} : 1 + 3 u^{2} + 3 u^{4} + u^{6}

= \int 1 + 3 u^{2} + 3 u^{4} + u^{6} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d u + \int 3 u^{2} d u + \int 3 u^{4} d u + \int u^{6} d u

\int 1 d u = u

\int 3 u^{2} d u = u^{3}

\int 3 u^{4} d u = \frac{3 u ^{5}}{5}

\int u^{6} d u = \frac{u ^{7}}{7}

= u + u^{3} + \frac{3 u ^{5}}{5} + \frac{u ^{7}}{7}

Substitute back

u = tan (x)

= tan (x) + tan^{3} (x) + \frac{3 tan ^{5} ( x )}{5} + \frac{tan ^{7} ( x )}{7}

Add a constant to the solution

= tan (x) + tan^{3} (x) + \frac{3 tan ^{5} ( x )}{5} + \frac{tan ^{7} ( x )}{7} + C

\frac{\partial}{\partial x} (40 x y \cdot e^{5 x^{2} - 2 y})

\int \frac{1}{4} x^{3} d x

\int (\frac{x - x ^{3} e ^{x} + x ^{2}}{x ^{3}}) d x

y^{^{'}} = \frac{1 - y ^{2}}{e ^{x}}

n = 0 \sum \infty p^{n}

What is the integral of sec^8(x) ?
The integral of sec^8(x) is tan(x)+tan^3(x)+(3tan^5(x))/5+(tan^7(x))/7+C